= Solution
With the stated time rescaling, $\dot x=\varepsilon^3u'$ and $\ddot x=\varepsilon^4u''$, where primes refer to $\widetilde t$. Dividing by $\varepsilon^4$ gives
$$
u''=u^2-\kappa-\varepsilon(\lambda+u)u'.
$$
At $\varepsilon=0$, put $v=u'$ and define the <Hamiltonian>
$$
\boxed{H(u,v)=\frac12v^2+\kappa u-\frac13u^3.}
$$
Its <orbital derivative> is $H'=v(u^2-\kappa)+(\kappa-u^2)v=0$, so this is a <conserved quantity>. The <Hamiltonian system> has $u'=H_v=v$, $v'=-H_u=u^2-\kappa$.
For $\kappa>0$, set $s=\sqrt\kappa$. The potential $U(u)=\kappa u-u^3/3$ has a minimum at $-s$ and a maximum at $s$. The corresponding <center equilibrium> and <saddle equilibrium> have energies $H_c=-2s^3/3$ and $H_s=2s^3/3$. Levels $H_c<H<H_s$ contain closed ovals around $(-s,0)$, as well as separate unbounded components on the right. At $H=H_s$, the center's ovals limit to a <homoclinic orbit> to the saddle, turning at $u=-2s$. Its equation is
$$
v^2=\frac23(u-s)^2(u+2s),\qquad -2s\leq u\leq s,
\qquad \boxed{H_{\mathrm{hom}}=\frac23\kappa^{3/2}.}
$$
The same energy also has external unbounded saddle separatrices for $u>s$. Levels above $H_s$ pass over the potential barrier and are open. Along the upper halves $u'=v>0$, so the closed loops are traversed clockwise. If $\kappa=0$, the two critical points coalesce into a degenerate <equilibrium>; if $\kappa<0$, the potential is strictly decreasing and there is no center or homoclinic loop.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-78-hamiltonian.png]
{title=Hamiltonian level curves for kappa equal to one, showing the center, saddle and homoclinic loop}
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